[数据结构与算法]算法数据结构（二十八）----二叉树的Morris遍历

Morris遍历

一种遍历二叉树的方式，并且时间复杂度O(N)，额外空间复杂度O(1)

通过利用原树中大量空闲指针的方式，达到节省空间的目的

?Morris遍历细节

假设来到当前节点cur，开始时cur来到头节点位置

1）如果cur没有左孩子，cur向右移动(cur = cur.right)

2）如果cur有左孩子，找到左子树上最右的节点mostRight：

? a.如果mostRight的右指针指向空，让其指向cur，

? 然后cur向左移动(cur = cur.left)

? b.如果mostRight的右指针指向cur，让其指向null，

? 然后cur向右移动(cur = cur.right)

3）cur为空时遍历停止

?Morris遍历实质

建立一种机制：

对于没有左子树的节点只到达一次，

对于有左子树的节点会到达两次

morris遍历时间复杂度依然是O(N)

??Morris遍历实现

public static class Node {
		public int value;
		Node left;
		Node right;

		public Node(int data) {
			this.value = data;
		}
	}

public static void morris(Node head) {
		if (head == null) {
			return;
		}
		Node cur = head;
		Node mostRight = null;
		while (cur != null) {
			mostRight = cur.left;
			if (mostRight != null) {
				while (mostRight.right != null && mostRight.right != cur) {
					mostRight = mostRight.right;
				}
				if (mostRight.right == null) {
					mostRight.right = cur;
					cur = cur.left;
					continue;
				} else {
					mostRight.right = null;
				}
			}
			cur = cur.right;
		}
	}

Morris遍历实现二叉树的先序遍历?

//第一次到达时，打印
	public static void morrisPre(Node head) {
		if (head == null) {
			return;
		}
		Node cur = head;
		Node mostRight = null;
		while (cur != null) {
			mostRight = cur.left;
			if (mostRight != null) {
				while (mostRight.right != null && mostRight.right != cur) {
					mostRight = mostRight.right;
				}
				if (mostRight.right == null) {
					System.out.print(cur.value + " ");
					mostRight.right = cur;
					cur = cur.left;
					continue;
				} else {
					mostRight.right = null;
				}
			} else {
				System.out.print(cur.value + " ");
			}
			cur = cur.right;
		}
		System.out.println();
	}

Morris遍历实现二叉树的中序遍历?

//只能到达一次的第一次打印，可以到达两次的第二次到达才打印
	public static void morrisIn(Node head) {
		if (head == null) {
			return;
		}
		Node cur = head;
		Node mostRight = null;
		while (cur != null) {
			mostRight = cur.left;
			if (mostRight != null) {
				while (mostRight.right != null && mostRight.right != cur) {
					mostRight = mostRight.right;
				}
				if (mostRight.right == null) {
					mostRight.right = cur;
					cur = cur.left;
					continue;
				} else {
					mostRight.right = null;
				}
			}
			System.out.print(cur.value + " ");
			cur = cur.right;
		}
		System.out.println();
	}

?Morris遍历实现二叉树的后序遍历?

//只有可以到达两次的节点，在第二次到达时，逆序打印该节点的右边界
	//整个树遍历后，逆序打印整棵树的右边界
	public static void morrisPos(Node head) {
		if (head == null) {
			return;
		}
		Node cur = head;
		Node mostRight = null;
		while (cur != null) {
			mostRight = cur.left;
			if (mostRight != null) {
				while (mostRight.right != null && mostRight.right != cur) {
					mostRight = mostRight.right;
				}
				if (mostRight.right == null) {
					mostRight.right = cur;
					cur = cur.left;
					continue;
				} else {
					mostRight.right = null;
					printEdge(cur.left);
				}
			}
			cur = cur.right;
		}
		printEdge(head);
		System.out.println();
	}

	//先链表反转，遍历后再反转回去
	public static void printEdge(Node head) {
		Node tail = reverseEdge(head);
		Node cur = tail;
		while (cur != null) {
			System.out.print(cur.value + " ");
			cur = cur.right;
		}
		reverseEdge(tail);
	}

	public static Node reverseEdge(Node from) {
		Node pre = null;
		Node next = null;
		while (from != null) {
			next = from.right;
			from.right = pre;
			pre = from;
			from = next;
		}
		return pre;
	}

题目一

给定一棵二叉树的头节点head

求以head为头的树中，最小深度是多少？

public static class Node {
		public int val;
		public Node left;
		public Node right;

		public Node(int x) {
			val = x;
		}
	}

	public static int minHeight1(Node head) {
		if (head == null) {
			return 0;
		}
		return p(head);
	}

	// 返回x为头的树，最小深度是多少
    // 使用递归求解
	public static int p(Node x) {
		if (x.left == null && x.right == null) {
			return 1;
		}
		// 左右子树起码有一个不为空
		int leftH = Integer.MAX_VALUE;
		if (x.left != null) {
			leftH = p(x.left);
		}
		int rightH = Integer.MAX_VALUE;
		if (x.right != null) {
			rightH = p(x.right);
		}
		return 1 + Math.min(leftH, rightH);
	}

?

	// 根据morris遍历改写
	//1.每到一个节点，可以知道它的level
	//2.每到一个节点，可以判断出是否为叶子节点
	public static int minHeight2(Node head) {
		if (head == null) {
			return 0;
		}
		Node cur = head;
		Node mostRight = null;
		int curLevel = 0;
		int minHeight = Integer.MAX_VALUE;
		while (cur != null) {
			mostRight = cur.left;
			if (mostRight != null) {
				int rightBoardSize = 1;
				while (mostRight.right != null && mostRight.right != cur) {
					rightBoardSize++;
					mostRight = mostRight.right;
				}
				if (mostRight.right == null) { // 第一次到达
					curLevel++;
					mostRight.right = cur;
					cur = cur.left;
					continue;
				} else { // 第二次到达
					if (mostRight.left == null) {
						minHeight = Math.min(minHeight, curLevel);
					}
					curLevel -= rightBoardSize;
					mostRight.right = null;
				}
			} else { // 只有一次到达
				curLevel++;
			}
			cur = cur.right;
		}
		int finalRight = 1;
		cur = head;
		while (cur.right != null) {
			finalRight++;
			cur = cur.right;
		}
		if (cur.left == null && cur.right == null) {
			minHeight = Math.min(minHeight, finalRight);
		}
		return minHeight;
	}